Question

In Fig.8.51, a block slides down an incline. As it moves from point Ato point B, which are 5.0 m apart, force $\overrightarrow{\mathbf{F}}$ acts on the block, with magnitude 2.0 N and directed down the incline. The magnitude of the frictional force acting on the block is 10 N . If the kinetic energy of the block increases by 35 J between Aand B, how much work is done on the block by the gravitational force as the block moves from Ato B?

Short answer

The amount of work done on the block by the gravitational force is$W_g=75\mathrm{J}$

Step-by-step solution

Step 1: Given Data

The distance between point A and point B, d = 5.0 m

The magnitude of applied force, F = 2.0 N

The magnitude of the friction force, f = 10 N

The increase in kinetic energy,$∆K=35\mathrm{J}$

Determining the concept

Use the equation of work relating mechanical energy and thermal energy. Mechanical energy is the sum of kinetic and potential energy. Thermal energy is the energy contained within a system that is responsible for its temperature.

Formulae are as follow:

$$\begin{gathered} W=∆E_{mec}+∆E_{th} \\ ∆E_{mec}=∆K+∆U \\ ∆E_{th}=fd \\ W=F.d \end{gathered}$$

where, $∆K$is change in kinetic energy,$∆U$ is change in potential energy, f, Fare forces,$∆E_{mec},∆E_{th}$ are mechanical and thermal energies, d is displacement and W is work done.

Determining the amount of work done on the block by the gravitational force

The equation for work done is,

$$\begin{gathered} W=∆E_{mec}+∆E_{th} \\ F.d=∆K+∆U+fd \end{gathered}$$

So,

$$\begin{gathered} F.d=∆K+∆U+fd \\ =2.0\times 5.0-35-10\times 5.0 \\ ∆U=-75\mathrm{J} \end{gathered}$$

The work done by gravitational energy will be equal to the change in gravitational potential energy. So,

$W_g=∆U=-75\mathrm{J}$

Hence,the amount of work done on the block by the gravitational force is$W_g=-75\mathrm{J}$.

Therefore, the work done by gravitation force can be found using the equation of work relating to energy.